RLC Resonance, Q Factor and Bandwidth Calculator

Every LC pair rings at one frequency; the resistance decides how enthusiastically. The same three numbers describe a resonant converter tank, an EMI filter pole, an unwanted layout resonance and a snubber problem - only which resistance counts differs. Series and parallel circuits share the frequency and characteristic impedance; the resistance enters upside down between them, which is the single most common RLC mistake.

Resonance response versus frequency for a high-Q and a low-Q circuit |H|ff0high Q: tall, narrow, slow to settlelow Q: short, wideBW = f0/Q
Q sets everything visible: the peak height, the -3 dB width f0/Q, and - though a frequency plot hides it - how many cycles the circuit rings after a disturbance, Q/pi. Tall and narrow is wanted in a tank, dreaded in a layout.

Series Q = Z0/R falls as R grows; parallel Q = R/Z0 rises. A resonant converter tank is series; a loaded filter output is parallel.

Inputs — enter your values

Solve for this from a target frequency: enter the frequency you want and choose "Solve for capacitance".

Series: total loop loss - winding ESR plus capacitor ESR plus trace. Parallel: the load across the tank.

Results — computed for you

sqrt(L/C) - what the tank current and voltage trade at. Peak resonant voltage is roughly Z0 times the drive current.

1/(2Q). Critically damped at 1 - a snubber design target. Below about 0.5 a step visibly rings.

The envelope of a disturbance decays as exp(-t/tau) with tau = 2Q/w0.

Q/pi. Count the ring cycles on a scope trace and multiply by pi to estimate Q directly.

Choosing something to solve for makes Resonant frequency an editable target and computes the chosen input from it.

Formula

f0 = 1 / (2 pi sqrt(L C)), Z0 = sqrt(L / C) series: Q = Z0 / R parallel: Q = R / Z0 zeta = 1/(2Q), BW = f0 / Q, tau = 2Q / w0
Z0 characteristic impedance of the tank (Ohm)
Q quality factor - energy stored over energy lost per radian
zeta damping ratio of the equivalent second-order system
tau ringing envelope decay time constant (s)

What this model assumes, and where it stops

Assumptions

  • Ideal linear L, C and R - no saturation, no bias dependence, no frequency dependence over the band of interest.
  • One lumped resistance. Real tanks lose in the winding, the capacitor and the core simultaneously; lump them into one series value at the operating frequency.
  • The -3 dB bandwidth and decay expressions assume Q above about 2; heavily damped circuits stop being "resonant" in any useful sense.
  • Second-order dynamics only - one L, one C.

Limitations

  • Component parasitics move real resonances: a capacitor is inductive above self-resonance and an inductor capacitive, so verify the parts are still themselves at f0.
  • Winding resistance rises with frequency through skin and proximity effect, so a Q computed from the DC resistance flatters the tank - sometimes by several times.
  • The half-power points sit geometrically around f0, not symmetrically; the bandwidth is exact but its centre shifts visibly below Q of about 5.
  • A loaded resonant converter tank sees an effective resistance set by the rectifier and load, not a physical resistor - derive it first (for an LLC, the fundamental-mode approximation), then use this page.

When you need a 3D field solution instead

Closed-form models like the one above hold on idealised geometry. These are the cases where they stop being good enough and a full 3D electromagnetic and thermal solution is the only way to get a trustworthy answer:

  • Self-resonance of a real wound component, where turn-to-turn capacitance is a 3D field quantity no lumped model predicts well.
  • Busbar and layout resonances in the tens of megahertz, set by loop geometry rather than intentional components.
  • Q of a tank at high frequency, where proximity-effect winding loss - a field problem - dominates the resistance.

Common questions

How do I measure Q on the bench?

Two easy ways. Count the visible ring cycles after a step or an injected pulse: the envelope falls to 1/e in Q/pi cycles, so ten visible cycles is a Q of roughly thirty. Or sweep and read the -3 dB width: Q = f0 divided by bandwidth. The two disagree when the resistance is frequency dependent - which for wound components above a few hundred kilohertz it always is.

Why does the resistance act oppositely in series and parallel circuits?

In a series loop the current passes through R every cycle, so bigger R burns more of the stored energy: Q = Z0/R. In a parallel tank the resistor sits across the swing, and a bigger R draws less current from it: Q = R/Z0. Mixing these up inverts the answer by Q^2 - with the default tank that is a factor of 400. When in doubt, convert everything to one topology first.

How do I size an RC snubber from this page?

Read the ringing frequency off the scope, estimate the parasitic L and C from f0 and Z0 (two equations, two unknowns: L = Z0/w0, C = 1/(Z0 w0)), then choose the snubber resistor near that Z0 - this lands the damping ratio near the critical value where ringing dies in about one cycle. The snubber capacitor is typically 3 to 5 times the parasitic C, trading damping speed against dissipation.

References

  • Erickson & Maksimovic - Fundamentals of Power Electronics, 3rd ed. - resonant conversion
  • Bowick - RF Circuit Design, 2nd ed. - resonant circuits, Q and loaded Q

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